小编Bac*_*ria的帖子

据我所知,在EJB 2.x中,客户端使用home接口请求对组件接口的引用,并使用该引用调用Enterprise java bean的业务方法.但是存根和骨架的概念对我来说并不清楚.

对组件接口的引用是否充当存根？然后哪一个充当骨架？

请澄清.

我试图从数据库中的对象做一个简单的加载,但我得到错误"无法初始化代理 - 没有会话",任何想法？谢谢

org.hibernate.LazyInitializationException: could not initialize proxy - no Session
    org.hibernate.proxy.AbstractLazyInitializer.initialize(AbstractLazyInitializer.jav a:167)
    org.hibernate.proxy.AbstractLazyInitializer.getImplementation(AbstractLazyInitializer.java:215)
    org.hibernate.proxy.pojo.javassist.JavassistLazyInitializer.invoke(JavassistLazyInitializer.java:190)
    com.myapp.domain.User_$$_javassist_0.getLogin(User_$$_javassist_0.java)
    com.myapp.validator.UserFormValidator.validate(UserFormValidator.java:34)

@Component 
public class UserFormValidator implements Validator {

@Autowired
private UserDAO userDAO;  

@Override
public boolean supports(Class<?> clazz) {
    return UserForm.class.equals(clazz);
}

public UserDAO getUserDAO() {
    return userDAO;
}

public void setUserDAO(UserDAO userDAO) {
    this.userDAO = userDAO;
}

@Override
public void validate(Object target, Errors errors) {
    User user = (User)getUserDAO().findById(new Integer(1));
    System.out.println ("User -> " + user.getLogin());
}
}

@Transactional
public class GenericDAOHibernateImpl <T, PK …

我正在尝试使用python在vtk中加载NIFTI图像.数据已经使用nibabel加载了,我可以使用pyplot对其进行可视化(每个绘图1个切片,我的数据是ndarray).但是我想加载它有一个ImagePlaneWidget ...就像这样.

我无法加载像imageplanewidget想要的数据(vtkDataSet)......我怎么能这样做？

import nibabel as nib
import matplotlib.pyplot as plt
import vtk

img = nib.load('image.nii.gz')
img_data = img.get_data()
print img_data.shape

def show_slices(slices):
    fig,axes = plt.subplots(1, len(slices))
    for i, slice in enumerate(slices):
        axes[i].imshow(slice.T, cmap="gray", origin="lower")

slice_0=img_data[100, :, :]
slice_1=img_data[:, 100, :]
slice_2=img_data[:, :, 100]
show_slices([slice_0, slice_1, slice_2])
plt.suptitle("Image")
plt.show()

plane = vtk.vtkImagePlaneWidget()
plane.SetInputData(img_data)

非常感谢,我是python和vtk的新手

我希望从具有 UUID 的节点/关系的查询返回一个最小值/最大值。我想将所有 UUID 收集为组，并为每组 UUID 找到最小值/最大值。

我试过

match (u:User) -[r:relation ]-> (n:Node) 
return (COLLECT (r.uuid)), max(n.value),  min(n.value)

我收到一个列表中的所有 UUID，并从所有 UUID 中收到每个分组的 UUID 的最小值和最大值。如何更改我的查询以获得所需的选择？

例子：

UUID  Value
1      1
1      5
1      7 
2      3
2      6

结果：1,2 1/7而不是1 1/7 - 2 3/6

我无法使用java配置获得条件弹簧批处理流程.我在春季批次样本,春季批次测试代码或堆栈溢出中看到的样本往往显示条件,其中单个步骤需要在条件上执行,或者它是最后一步,或两者.那不是我需要解决的情况.

在程序伪代码中,我希望它表现得像

initStep()
if decision1()
    subflow1()
middleStep()
if decision2()
    subflow2()
lastStep()

因此,subflow1和2是有条件的,但init,middle和last总是执行.这是我的剥离测试用例.在当前配置中,它只是在执行subflow1后退出.

public class FlowJobTest {

private JobBuilderFactory jobBuilderFactory;
private JobRepository jobRepository;
private JobExecution execution;

@BeforeMethod
public void setUp() throws Exception {
    jobRepository = new MapJobRepositoryFactoryBean().getObject();
    jobBuilderFactory = new JobBuilderFactory(jobRepository);
    execution = jobRepository.createJobExecution("flow", new JobParameters());
}

@Test
public void figureOutFlowJobs() throws Exception {

    JobExecutionDecider subflow1Decider = decider(true);
    JobExecutionDecider subflow2Decider = decider(false);

    Flow subflow1 = new FlowBuilder<Flow>("subflow-1").start(echo("subflow-1-Step-1")).next(echo("subflow-1-Step-2")).end();
    Flow subflow2 = new FlowBuilder<Flow>("subflow-2").start(echo("subflow-2-Step-1")).next(echo("subflow-2-Step-2")).end();

    Job job = jobBuilderFactory.get("testJob")
            .start(echo("init"))

            .next(subflow1Decider)
                .on("YES").to(subflow1)
            .from(subflow1Decider)
                .on("*").to(echo("middle")) …

试图了解super()方法何时被调用.在下面的代码中,Child类有一个带有this()的无参数构造函数,因此编译器无法插入super().那么如何调用父构造函数.

public class Parent
{
    public Parent()
    {
        System.out.println("In parent constructor");
    }
 }


public class Child extends Parent
{
private int age;

public Child()
{   
    this(10);
    System.out.println("In child constructor with no argument");
}

public Child(int age)
{
    this.age = age;
    System.out.println("In child constructor with argument");
}

public static void main(String[] args)
{
    System.out.println("In main method");
    Child child = new Child();
}

}

输出:

In main method

In parent constructor

In child constructor with argument

In child constructor with no argument

如何用Hibernate验证传入的参数？

在XML中

<bean id="validator" class="org.springframework.validation.beanvalidation.LocalValidatorFactoryBean"/>
<bean class="org.springframework.validation.beanvalidation.MethodValidationPostProcessor">
    <property name="validator" ref="validator"/>
</bean>

在Java中

 @Validated
 public class UserService

 @Override
 @NotNull
 public User registerUser(@NotEmpty String name, String username, String password, boolean google, boolean facebook)

这种方法不起作用我称之为错误参数的方法,验证不起作用.

我正在做一个需要重复配置文件的应用程序.但他们的商业模式包括1个试用期和2个常规期,一个是几天,另一个是每月无限期.

{  
   "name":"Regular Membership Billing Plan",
   "description":"This plan is for regular membership\r\n7 days trial\r\nBilling 23 days after trial\r\ninfinite montly",
   "type":"INFINITE",
   "payment_definitions":[  
      {  
         "name":"7 days trial regular product",
         "type":"TRIAL",
         "frequency":"WEEK",
         "frequency_interval":"1",
         "cycles":"1",
         "amount":{  
            "value":"2",
            "currency":"USD"
         }
      },
      {  
         "name":"23 days after trial regular product",
         "type":"REGULAR",
         "frequency":"DAY",
         "frequency_interval":"23",
         "cycles":"1",
         "amount":{  
            "value":"5",
            "currency":"USD"
         }
      },
      {  
         "name":"Monthly regular product",
         "type":"REGULAR",
         "frequency":"MONTH",
         "frequency_interval":"1",
         "cycles":"0",
         "amount":{  
            "value":"7.99",
            "currency":"USD"
         }
      }
   ]
}

我从响应中收到此错误:

"Payment definitions are invalid. Valid parameters are REGULAR or combination of TRIAL …

在我启动Hibernate 5应用程序的那一刻,我发现这些警告就像疯了一样.我怎么能摆脱他们？这是我的配置文件:

<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE hibernate-configuration PUBLIC "-//Hibernate/Hibernate Configuration DTD//EN" "http://hibernate.sourceforge.net/hibernate-configuration-3.0.dtd">
<!--
Program developed by Hassan Althaf.
Copyright © 2015, Hassan Althaf.
Website: http://hassanalthaf.com
-->

<hibernate-configuration>
    <session-factory>
        <property name="hibernate.dialect">
            org.hibernate.dialect.MySQLDialect
        </property>
        <property name="hibernate.connecton.driver_class">
            com.mysql.jdbc.Driver
        </property>
        <property name="hibernate.connection.url">
            jdbc:mysql://127.0.0.1:3306/telemart
        </property>
        <property name="hibernate.connection.username">
            root
        </property>
        <property name="hibernate.connection.password">
            php123
        </property>
        <mapping resource="com/hassanalthaf/telemart/users/User.hbm.xml"/>
        <mapping resource="com/hassanalthaf/telemart/customers/Customer.hbm.xml"/>
        <mapping resource="com/hassanalthaf/telemart/inventory/Product.hbm.xml"/>
        <mapping resource="com/hassanalthaf/telemart/orders/Order.hbm.xml"/>
        <mapping resource="com/hassanalthaf/telemart/orders/OrderItem.hbm.xml"/>
    </session-factory>
</hibernate-configuration>

这是整个输出:

ant -f /Users/hassan/NetBeansProjects/TeleMart/TeleMart jfxsa-run
init:
Deleting: /Users/hassan/NetBeansProjects/TeleMart/TeleMart/build/built-jar.properties
deps-jar:
Updating property file: /Users/hassan/NetBeansProjects/TeleMart/TeleMart/build/built-jar.properties
Duplicated project name in import. Project jfx-impl …

我正在编写junit测试,我需要从中读取文件src/test/resources

我的代码中有以下代码src/test/java/some_packages/UserTests:

InputStream inputStream = new FileInputStream(new File("abc.txt"));

该文件当然存在,但是当我运行我的测试时,我得到了 java.io.FileNotFoundException: abc.txt (No such file or directory)

我将资源文件夹标记为intellij中的测试资源